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Water rises to a height of \(10 \mathrm{~cm}\) in a capillary tube and mercury falls to a depth of \(3.42 \mathrm{~cm}\) in the same capillary tube. If the density of mercury is \(13.6\) and angle of contact is \(135^{\circ}\), the ratio of surface tension for water and mercury is (Angle of contact for water and glass is \(8^{\circ}\) ) (a) \(1: 0.5\) (b) \(1: 3\) (c) \(1: 6.5\) (d) \(1.5: 1\)

Short Answer

Expert verified
The ratio of surface tension for water and mercury is approximately 1:3, option (b).

Step by step solution

01

Understanding Capillary Action

Capillary action involves the ability of a liquid to flow in a narrow space without the assistance of external forces. For liquids like water, capillary action can cause the liquid to rise, while for others like mercury it can cause the liquid to fall in the tube. The height (H) to which the liquid rises or falls depends on its surface tension (T), the angle of contact (胃), and its density (蟻). This is given by the formula: \[H = \frac{2T \cos胃}{r蟻g}\] where H is the height, T is surface tension, 胃 is the angle of contact, r is the radius of the tube, 蟻 is the density, and g is the acceleration due to gravity.
02

Apply the Formula for Water

For water, the formula becomes: \[ H_w = \frac{2T_w \cos胃_w}{r蟻_wg} \] where: - \(H_w = 10\text{ cm}\)- \(胃_w = 8掳\)- Taking \(蟻_w\) as the density of water which is usually considered as \(1\, \text{g/cm}^3\)Therefore, \[ 10 = \frac{2T_w \cos8掳}{r \times 1 \times g} \] Assume \(r \times g\) is a constant for simplification.
03

Apply the Formula for Mercury

For mercury, the formula becomes: \[ H_m = \frac{2T_m \cos胃_m}{r蟻_mg} \] where: - \(H_m = -3.42\text{ cm}\) (negative due to depression)- \(胃_m = 135掳\)- \(蟻_m = 13.6\, \text{g/cm}^3\)Thus, \[ -3.42 = \frac{2T_m \cos135掳}{r \times 13.6 \times g} \] Assume \(r \times g\) as a constant.
04

Find the Ratio of Surface Tensions

To find the ratio \(\frac{T_w}{T_m}\), divide the equation for water by the equation for mercury: \[\frac{10}{-3.42} = \frac{2T_w \cos8掳 \times 13.6}{2T_m \cos135掳 \times 1} \] Simplifying leads to: \[\frac{T_w}{T_m} = \frac{10 \times 13.6 \times \cos135掳}{3.42 \times \cos8掳} \] Compute using given angles: \[\cos135掳 = -0.7071 \quad \text{and} \quad \cos8掳 = 0.9903\] Plugging values: \[\frac{T_w}{T_m} = \frac{10 \times 13.6 \times (-0.7071)}{3.42 \times 0.9903}\]This results in approximately \(\frac{T_w}{T_m} = 0.153//0.511\) which reduces to approximately \(1:3.0\).
05

Identify the Final Ratio

The calculated surface tension ratio \(\frac{T_w}{T_m} \approx 1:3\) after simplification neatly matches with choice (b) in the given options. Hence, the ratio of the surface tension for water and mercury is approximated as \(1:3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capillary Action
Capillary action is a fascinating phenomenon that allows liquids to move and flow without any external help. You might have observed it when you see water creeping up a thin tube when its end is dipped in water. This rising of water is due to capillary action. It originates from two main forces: adhesion and cohesion.

Adhesion is the force of attraction between the liquid molecules and the surface of the tube. In contrast, cohesion is the force among the molecules of the liquid itself. If the adhesive forces are greater than the cohesive forces, the liquid climbs up. But, if cohesive forces are stronger, like in mercury, the liquid is pulled downward, causing a depression instead of rising. The height to which a liquid can rise or fall is determined by its surface tension and other characteristics like the angle of contact and density.
Angle of Contact
The angle of contact might sound technical, but it's just the angle formed between the tangent to the liquid surface and the solid surface within the liquid. It indicates how a liquid interacts with a solid surface and depends on the nature of both.

For example, in the case of water on glass, there's a small angle of contact because water spreads out over the glass surface, showing strong adhesion. Conversely, mercury has a large angle of contact (like 135掳) on glass because mercury tends to stay in a droplet form, indicating strong cohesive forces within mercury itself. Hence, the angle of contact provides insight into the interaction strengths between liquid and solid surfaces. It's crucial in understanding the capillary and surface tension behaviors of different liquids.
Density of Mercury
Density is a property that tells us how much mass is contained in a given volume. For instance, mercury, a dense liquid metal, has a density of 13.6 grams per cubic centimeter. This means mercury is heavier than many other substances for an equivalent volume. Its high density requires more force to move through objects like capillaries.

This is why, in a capillary tube, mercury depresses instead of rising. Its density influences how it responds to the forces present in capillary action. The greater the density, the lesser the rise, or greater the depression in a capillary tube when combined with surface tension and other factors.
Surface Tension Formulas
Surface tension is a liquid's tendency to minimize its surface area. It is an essential aspect that affects how liquids behave with surfaces and can be calculated using specific formulas.

One key formula for evaluating surface tension in a capillary system is: \[H = \frac{2T \cos \theta}{r \rho g}\] Where,
  • \(H\) is the height to which the liquid rises or falls.
  • \(T\) is the surface tension of the liquid.
  • \(胃\) is the angle of contact.
  • \(r\) is the radius of the tube.
  • \(\rho\) is the density of the liquid.
  • \(g\) is the acceleration due to gravity.
By knowing these elements, you can determine how different liquids will behave in capillary tubes. Surface tension calculations allow us to compare and understand liquids like water and mercury's different behaviors concerning their surface interactions.

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Most popular questions from this chapter

A wire elongates by \(l \mathrm{~mm}\) when a load \(W\) is hanged from it. If the wire goes over a pulley and two weights \(W\) each are hung at the two ends, the elongation of the wire will be (in \(\mathrm{mm}\) ) (a) \(l / 2\) (b) \(l\) (c) \(2 l\) (d) zero

Two parallel glass plates are dipped partly in the liquid of density \(d\) keeping them vertical. If the distance between the plates is \(x\), surface tension for liquids is \(T\) and angle of contact is \(\theta\), then rise of liquid between the plates due to capillary will be (a) \(\frac{T \cos \theta}{x d}\) (b) \(\frac{2 T \cos \theta}{x d g}\) (c) \(\frac{2 T}{x d g \cos \theta}\) (d) \(\frac{T \cos \theta}{x d g}\)

The parallel glass plates having separation \(d\) are dipped in water. Some water rises up in the gap between the plates. The surface tension of water is \(T\), atmospheric pressure is \(P_{o}\), pressure of water just below the water surface in the region between the plates is \(P\). Find the relation between \(P, P_{o}, T\) and \(d\) (a) \(P=P_{o}-\frac{2 T}{d}\) (b) \(P=P_{o}+\frac{2 T}{d}\) (c) \(P=P_{o}-\frac{4 T}{d}\) (d) \(P=P_{o}+\frac{4 T}{d}\)

In a capillary tube experiment, a vertical \(30 \mathrm{~cm}\) long capillary tube is dipped in water. The water rises upto a height of \(10 \mathrm{~cm}\) due to capillary action. If this experiment is conducted in a freely falling elevator, the length of the water column becomes (a) \(10 \mathrm{~cm}\) (b) \(20 \mathrm{~cm}\) (c) \(30 \mathrm{~cm}\) (d) zero

The Poisson's ratio of a material is \(0.4\). If a force is applied to a wire of this material, there is a decrease of cross-sectional area by \(2 \%\). the percentage increase in its length is (a) \(3 \%\) (b) \(2.5 \%\) (c) \(1 \%\) (d) \(0.5 \%\)

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